Differential Equations

The order of a differential equation

The order of a differential equation is simply the order of the highest derivative appearing in it. No need to solve anything first.

What is the order of $2x^2\dfrac{d^2y}{dx^2} - 3\dfrac{dy}{dx} + y = 0$ ?

Solution: The highest derivative present is $\dfrac{d^2y}{dx^2}$ , a second derivative, so the order is 2.

Order and degree

The degree is the power of the highest-order derivative, once the equation is written as a polynomial in its derivatives. If a derivative sits inside a function like $\sin$ or $e^x$ , the degree is not defined.

Order is the highest derivative; degree is that derivative's power

Equation	Order	Degree
$\dfrac{dy}{dx} + y = x$	1	1
$\left(\dfrac{d^2y}{dx^2}\right)^3 + \left(\dfrac{dy}{dx}\right)^4 + y = 0$	2	3
$\left(\dfrac{dy}{dx}\right)^2 = 1 + x$	1	2
$\dfrac{d^2y}{dx^2} + \sin\left(\dfrac{dy}{dx}\right) = 0$	2	not defined

In the second row, the degree is 3 even though a 4th power appears. Degree belongs to the highest-order derivative, $\dfrac{d^2y}{dx^2}$ , not to the biggest exponent.

Three types of first-order equation

How to recognise and solve each type

Type	Form	Method
Variable separable	$\dfrac{dy}{dx} = g(x)\,h(y)$	Separate: $\dfrac{dy}{h(y)} = g(x)\,dx$ , then integrate both sides
Homogeneous	$\dfrac{dy}{dx} = F\left(\dfrac{y}{x}\right)$	Substitute $y = vx$ , so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$ , then separate
Linear	$\dfrac{dy}{dx} + P(x)\,y = Q(x)$	Multiply by the integrating factor $e^{\int P\,dx}$ : $y\cdot\text{IF} = \int Q\cdot\text{IF}\,dx$

The worked example below on separating variables covers the first type. The other two follow it.

Worked example: forming an equation from a family of curves

Form the differential equation representing the family of curves $y = mx$ , where $m$ is an arbitrary constant.

Solution: Differentiate $y = mx$ with respect to $x$ :

\dfrac{dy}{dx} = m

Since $m = \dfrac{y}{x}$ from the original equation, substituting eliminates the arbitrary constant, giving the differential equation:

\dfrac{dy}{dx} = \dfrac{y}{x}

Worked example: solving by separating variables

Find the equation of the curve passing through $(-2, 3)$ , given that the slope of the tangent at any point $(x, y)$ is $\dfrac{2x}{y^2}$ .

Solution: The slope is $\dfrac{dy}{dx} = \dfrac{2x}{y^2}$ . Separating variables and integrating both sides:

y^2\,dy = 2x\,dx \;\Rightarrow\; \int y^2\,dy = \int 2x\,dx \;\Rightarrow\; \dfrac{y^3}{3} = x^2 + C

Using the point $(-2, 3)$ to find $C$ :

\dfrac{27}{3} = 4 + C \;\Rightarrow\; 9 = 4 + C \;\Rightarrow\; C = 5

So $\dfrac{y^3}{3} = x^2 + 5$ , or equivalently:

y^3 = 3x^2 + 15

Worked example: a homogeneous equation

Solve $\dfrac{dy}{dx} = \dfrac{x + y}{x}$ for $x > 0$ .

Solution: The right side is $1 + \dfrac{y}{x}$ , a function of $\dfrac{y}{x}$ , so the equation is homogeneous. Substitute $y = vx$ :

v + x\dfrac{dv}{dx} = 1 + v \;\Rightarrow\; x\dfrac{dv}{dx} = 1 \;\Rightarrow\; \int dv = \int \dfrac{dx}{x} \;\Rightarrow\; v = \ln x + C

Replacing $v$ with $\dfrac{y}{x}$ :

y = x\ln x + Cx

Worked example: a linear equation

Solve $\dfrac{dy}{dx} + \dfrac{y}{x} = x^2$ for $x > 0$ .

Solution: This is linear with $P(x) = \dfrac{1}{x}$ and $Q(x) = x^2$ . The integrating factor is $e^{\int \frac{1}{x}dx} = e^{\ln x} = x$ . Multiplying through by it turns the left side into the derivative of $xy$ :

\dfrac{d}{dx}(xy) = x^3 \;\Rightarrow\; xy = \dfrac{x^4}{4} + C \;\Rightarrow\; y = \dfrac{x^3}{4} + \dfrac{C}{x}

Common mistakes

Confusing order and degree. Order is which derivative is highest; degree is the power it is raised to.
Giving a degree when it isn't defined. If a derivative appears inside $\sin$ , $\log$ or an exponential, there is no degree.
Dropping $C$ before dividing. In $xy = \dfrac{x^4}{4} + C$ , dividing by $x$ gives $\dfrac{C}{x}$ , not $C$ . Add the constant as soon as you integrate.
Writing the integrating factor wrongly. It is $e^{\int P\,dx}$ , using $P$ from the standard form with the coefficient of $\dfrac{dy}{dx}$ equal to 1. Divide through first if it isn't.